Declare array in bash/shell

Question

I am confused about how to declare an array in bash.

Presume:

$arr[1]=x
$arr[2]=y
$arr[3]=z

Code:

declare -A/-a object
fo var in ${arr[@]}
do
    object["a"]="${arr[1]}"
    object["b"]="${arr[2]}"
    object["c"]="${arr[3]}"
    echo ${object["a"]}
done

Output: z

What I want is :x

If I commented object["b"] and object["c"], then the output is correct：x

No matter -A or -a, still does not work. is this problem related to my bash environment?

GNU bash,version 3.2.57(1) -release (x86_64_apple-darwin18)

Put a valid shebang on your code and paste it at https://shellcheck.net for validation.recommendation. — Jetchisel, Sep 09 '21 at 06:41
Upper case `-A` to declare an associative array (Has to be done before use, unlike regular ones) — Shawn, Sep 09 '21 at 06:50
I saw some duplicates here in S.O. so I deleted my answer twice :-) — Jetchisel, Sep 09 '21 at 07:16

score 0 · Answer 1 · answered Sep 09 '21 at 07:14

0

The indexed array requires the subscript must be evaluated to a number. In this case, object["a"] object["b"] object["c"] is treated as object[0] object[0] object[0] Associate array can be used if the key is an arbitrary string. To declare it using -A instead of -a

-a Each name is an indexed array variable (see Arrays above).

-A Each name is an associative array variable (see Arrays above).

answered Sep 09 '21 at 07:14

nhatnq

1,173
7
16

not worked, please see my update – YYZZYY Sep 09 '21 at 20:51

Declare array in bash/shell

1 Answers1